20172325 "Java program design" the first week to summarize the learning contents of teaching materials to summarize the problems in learning and solving process
Textbook learning Problems first go to https://shimo.im/doc/1i1gldfsojIFH8Ip/to see, if others did not ask the same question, you can edit the document to add, and then copy their questions to the following:
- Question 1:xxxxxx
- Issue 1 Solution: XXXXXX
- Question 2:xxxxxx
- Issue 2 Solution: XXXXXX
- ...
Textbook Exercise Solutions
- EX2.1 What is the order of the following growth functions?
- a.10n^2+100n+1000
- Solution: O (n^2)
- B.10n^3-7
- Solution: O (n^3)
- C.2^n+100n^3
- Solution: O (2^n)
- D.n^2 log2 (N)
- Solution: O (n^2 log2 (n))
- EX2.4 Please determine the growth function and order of the following code snippet:
for(int count = 0 ; count < n ; count++) for(int count2 = 0 ; count2 < n ; count2 = count2 + 2) { System.out.println(count,count2); }}
- 解:嵌套循环,内层循环的循环次数是n/2,外层循环的循环次数是n,所以增长函数为:F(n)=(n^2)/2,阶次为O(n^2)。
- EX 2.5: Please determine the growth function and order of the following code snippet:
for(int count = 0 ; count < n ; count++) for(int count2 = 1 ; count2 < n ; count2 = count2 * 2) { System.out.println(count,count2); }}
- 解:嵌套循环,内层循环的循环次数是logn,外层循环的循环次数是n,所以增长函数为:F(n)=nlog2(n),阶次为O(n·log2(n))。
Code Hosting
No code this week
Pair and peer review templates:
- Blogs that are worth learning or questions:
- Something worth learning or doing in your code:
- Based on the scoring criteria, I scored this blog: xx points. The score is as follows: XXX
reviewed the classmates blog and code
- This week's study of the knot
- 20172306
- Pair of photos
- Pairs of learning content
Other (sentiment, thinking, etc., optional)
Xxx
Xxx
Learning progress Bar
|
lines of code (new/cumulative) |
Blog Volume (Add/accumulate) |
Learning Time (new/cumulative) |
Goal |
5000 rows |
30 Articles |
400 hours |
First week |
0/0 |
1/1 |
8/8 |
Resources
20172325 Java Programming the first week of learning summary